How is $\sin 90° = 1$ possible?

Consider polar coordinates, $(r \cos\theta, r \sin \theta)$ in a unit circle such that $r=1$. Then, we can see that, the mapping in the first quadrant inside the unit circle is just $(\cos \theta, \sin \theta)$. Now, consider a moving point $A$ starting from $\theta=0°$ to $\theta=90°$ on the circumference of the circle. Now, $OA$ is the hypotenuse of the right angle triangle inside the circle. Now, it is clear that at $\theta=90°$, the hypotenuse and the perpendicular are the same line in the $x$-axis (i.e., they coincide). So, $\sin(90°)=\frac{p}{h}=1$. As per your argument, the case is two sides coinciding rather than being parallel, because by Pythagoras theorem, we have $h^2=p^2+b^2$, so if $p$ increases then $b$ must decrease, and $h=p$ iff $b=0$.

Where, h is the radius of the unit circle, p is the perpendicular drawn from (aka height) the point in the circumference to the x-axis, and b is the distance of intersection of p and x-axis from the origin.

You can very well admit the concept of flat triangles (coming in two flavors, with a $0°$ and a $180°$ angle).

Anyway, the usual definition of the sine does not involve a triangle. Rather, the trigonometrical circle and the projection of a point at a given angle to the vertical axis. Indisputably, $\sin90°=1$. Simlarly, $\cos90°=0$.

A trivial Pythagorean triple, generated by Euclid's formula where $m=n$ such as $(A,B,C)=(0,2,2)$, is a vertical line $(A=0)$ on the $y$ axis. Since $\frac{C}{B}=1, \sin90^\circ=1$.

As noted in a comment, this answer to a previous similar question may well help.

To answer your question directly: I offer that mathematicians made $\sin(x)$ “well defined” for values of $x$ that are smaller than $0^{\circ}$ or larger than $90^{\circ}$ (that is, outside the range of values that are valid for an interior angle of a triangle) when it became useful to do so. “Well defined” is a technical term that means, roughly, that everyone agrees on the answer.

As a simple example, imagine an object that is moving in the plane at constant velocity. The velocity can be described as a speed $v$ and an angle $\theta$ measured clockwise from the $x$ axis. Now suppose that we ask the question, “How fast is the object moving in the $y$ direction?”

If $\theta$ is less than $90^{\circ}$ then trigonometry gives the answer $v \sin(\theta)$. For values of $\theta$ equal or larger than $90^{\circ}$, it’s possible to calculate an answer by deducting an appropriate multiple of $90^{\circ}$ to bring $\theta$ back into the range of valid triangle interior angles. But that’s tedious. The answer “should” be $v \sin(\theta)$ for any direction $\theta$. To make this answer work, the definition of $\sin$ has to be extended from the range of valid triangle interior angles $(0^{\circ},90^{\circ})$ to the range of valid directions $[0^{\circ},360^{\circ})$. In this extension definition, $\sin(90^{\circ})=1$.